inches. In each postcard, the length is 1 times the width
Which of the following could be the dimensions of the postcard
or one of the enlargements? Mark all that apply.
Answer:
The Answer is: There are 5 large boxes and 8 small boxes. (The graphical plot will need to be solved using the software).
Step-by-step explanation:
Let xx = small boxes and yy = large boxes:
xx + yy = 13
xx = 13 - yy
The number of small boxes times 20, plus the number of large boxes times 40 is equal to 360:
20xx + 40yy = 360
Substitute the algebraic value of xx, so we only have one variable to consider and solve for yy:
20(13 - yy) + 40yy = 360
260 - 20yy + 40yy = 360
260 + 20yy = 360
20yy = 100
yy = 100 / 20 = 5 large boxes.
Solve for xx:
xx = 13 - yy
xx = 13 - 5 = 8 small boxes.
There are 5 large boxes and 8 small boxes. (The graphical plot will need to be solved using the software).
Proof:
20xx + 40yy = 360
20(8) + 40(5) = 360
160 + 200 = 360
360 = 360
To solve the problem graphically, set up a system of equations: x + y = 13 and 20x + 40y = 360. Graph the equations to find the intersection point which represents the solution.
To solve this problem graphically, we can set up a system of equations based on the given information. Let's denote the number of small boxes as x and the number of large boxes as y. From the given information, we can write the following equations:
x + y = 13 (equation 1, representing the total number of boxes)
20x + 40y = 360 (equation 2, representing the total number of books)
To graph these equations, we can plot them on a coordinate plane, where the x-axis represents the number of small boxes (x) and the y-axis represents the number of large boxes (y). The intersection point of these two lines will represent the solution to the system of equations, giving us the values of x and y (the number of small and large boxes sent, respectively).
#SPJ2
Answer:
You started with 150.